TITLE: lifting property of universal cover of $\mathbb{RP}^n$
QUESTION [6 upvotes]: I have a question. If we have a map $f:\mathbb RP^n\rightarrow \mathbb RP^n$, then can we always lift it to a map $g:S^n\rightarrow S^n$ such that the diagram commutes?
$$\begin{array}
$S^n & \stackrel{g}{\longrightarrow} & S^n\\
\downarrow{p} & & \downarrow{p} \\
\mathbb RP^n & \stackrel{f}{\longrightarrow} & \mathbb RP^n  
\end{array}
$$
Here $S^n$ is the universal cover of $\mathbb RP^n$. I have another question. Does any map $h: Y\rightarrow X$ can always be lifted to $\hat h: Y\rightarrow\hat X $ where $\hat X$ is the universal cover of X? 
Any help will be appreciated!

REPLY [5 votes]: In general, the lifting property says that given a map $h:Y\to X$, and a covering map $p: X' \to X$, the map $h$ lifts to a map $\hat h: Y\to X'$ (with designated base point $y_0$ mapping to designated base point $x_0'$ above $x_0=h(y_0)$) if and only if the containment $f(\pi_1(Y,y_0))\subseteq p(\pi_1(X',x_0'))$ is satisfied. In particular, if $Y$ is simply connected then this will always be true, so the answer to your first question is yes. On the other hand, the answer to the second question is no: a simple counterexample is given by $Y=X=S^1$ with $h$ being the identity map; if $h$ lifted to a map $\hat h: S^1 \to \mathbb R$, then $\hat h$ would be null-homotopic, and hence $h=p \circ \hat h$ would be null-homotopic as well, a contradiction, since we know that in fact $h$ is not null-homotopic.